A random sample of 90 observations produced a mean x=25.6 and a standard deviati

Question

A random sample of 90 observations produced a mean x=25.6 and a standard deviation s = 2.6. a. Find a 95% confidence interval for b. Find a 90% confidence interval for . c. Find a 99% confid . ence interval for a. The 95% confidence interval is (LI (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.) b. The 90% confidence interval is 1]LD (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.) C. The 99% confidence interval is (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

Explanation / Answer

Mean is 25.6 and s is 2.6, for sample size of 90, the standard error is SE/sqrt(N)=2.6/sqrt(90)=0.2741

a) for 95% confidence, the z= 1.96, thus the lower bound is mean-SE*z =25.6-1.96*0.2741=25.06.  Upper bound is mean+SE*z =25.6+1.96*0.2741=26.14

b) for 90% confidence, the z=1.65, thus the lower bound is mean-SE*z=25.6-1.65*0.2741=25.15. upper bound is mean+SE*z=25.6+1.65*0.2741=26.05

c) for 99% confidence, the z=2.576, thus lower bound is mean-SE*z=25.6-2.576*0.2741=24.89 and upper bound is mean+SE*z=25.6+2.576*0.2741=26.31

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